Skillnad mellan versioner av "1.5a Lösning 6b"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| (8 mellanliggande versioner av samma användare visas inte) | |||
| Rad 1: | Rad 1: | ||
| − | <math>\begin{align} {1+x \over 1} & = {1 \over x} \qquad\qquad | \ | + | <math>\begin{align} {1+x \over 1} & = {1 \over x} \qquad\qquad | \, \cdot x \\ |
x \cdot (1 + x) & = 1 \\ | x \cdot (1 + x) & = 1 \\ | ||
| − | x + x^2 & = 1 \qquad\qquad | \ | + | x + x^2 & = 1 \qquad\qquad | \, -1 \\ |
x^2 + x - 1 & = 0 \\ | x^2 + x - 1 & = 0 \\ | ||
x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ | x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ | ||
x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ | x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ | ||
x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots \\ | x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots \\ | ||
| − | x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-{1+\sqrt{5} \over 2} \; {\rm :negativ} | + | x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-\,{1+\sqrt{5} \over 2}\; {\rm :negativ} \\ |
| − | g & = {\sqrt{5}-1 \over 2} \approx | + | g & = {\sqrt{5}-1 \over 2} \approx 0,618033989 |
\end{align}</math> | \end{align}</math> | ||
Nuvarande version från 8 januari 2019 kl. 16.47
\(\begin{align} {1+x \over 1} & = {1 \over x} \qquad\qquad | \, \cdot x \\ x \cdot (1 + x) & = 1 \\ x + x^2 & = 1 \qquad\qquad | \, -1 \\ x^2 + x - 1 & = 0 \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots \\ x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-\,{1+\sqrt{5} \over 2}\; {\rm :negativ} \\ g & = {\sqrt{5}-1 \over 2} \approx 0,618033989 \end{align}\)
